Question: Rewrite the function by completing the square. $h(x)= 4 x^{2} +4 x +1$ $h(x)=$
$\begin{aligned} h(x)&=4 x^2 +4 x +1 \\\\ &=4 \left(x^2 + x\right) +1 \end{aligned}$ Now we want to complete $x^2 +x$ into a perfect square. To do that, we should add $\left(\dfrac{{+1}}{2}\right)^2={\dfrac{1}{4}}$ to it: $x^2{+}x+{\dfrac{1}{4}}=\left(x +\dfrac{1}{2}\right)^2$ We add ${\dfrac{1}{4}}$ inside the parentheses, and subtract ${4}\cdot{\dfrac{1}{4}}$ outside them, to keep the expression equivalent. $\begin{aligned} &\phantom{=}{4} \left(x^2 + x\right) +1 \\\\ &={4}\left(x^2 + x+{\dfrac{1}{4}}\right) +1 -{4}\cdot{\dfrac{1}{4}} \\\\ &=4 \left(x +\dfrac{1}{2}\right)^2 +1 -1 \\\\ &=4 \left(x +\dfrac{1}{2}\right)^2 +0 \end{aligned}$ In conclusion, the function after completing the square is written as: $h(x)=4 \left(x +\dfrac{1}{2}\right)^2 +0$